class Solution {
    // blank 返回长度为 n 的由空格组成的字符串
    string blank(int n) {
        return string(n, ' ');
    }

    // join 返回用 sep 拼接 [left, right) 范围内的 words 组成的字符串
    string join(vector<string> &words, int left, int right, string sep) {
        string s = words[left];
        for (int i = left + 1; i < right; ++i) {
            s += sep + words[i];
        }
        return s;
    }

public:
    vector<string> fullJustify(vector<string> &words, int maxWidth) {
        vector<string> ans;
        int right = 0, n = words.size();
        while (true) {
            int left = right; 
            int sumLen = 0; 
            while (right < n && sumLen + words[right].length() + (right - left) <= maxWidth) {
                sumLen += words[right].length();
                right++;
            }

            // 当前行是最后一行：单词左对齐，且单词之间应只有一个空格，在行末填充剩余空格
            if (right == n) {
                string s = join(words, left, n, " ");
                ans.emplace_back(s + blank(maxWidth - s.length()));
                return ans;
            }

            int numWords = right - left;
            int numSpaces = maxWidth - sumLen;

            // 当前行只有一个单词：该单词左对齐，在行末填充剩余空格
            if (numWords == 1) {
                ans.emplace_back(words[left] + blank(numSpaces));
                continue;
            }

            // 当前行不只一个单词
            int avgSpaces = numSpaces / (numWords - 1);
            int extraSpaces = numSpaces % (numWords - 1);
            string s1 = join(words, left, left + extraSpaces + 1, blank(avgSpaces + 1)); // 拼接额外加一个空格的单词
            string s2 = join(words, left + extraSpaces + 1, right, blank(avgSpaces)); // 拼接其余单词
            ans.emplace_back(s1 + blank(avgSpaces) + s2);
        }
    }
};